SPUG: range operator

[Florentin]

Thank you everybody for explanations  and interpreting perldoc pages.

Florentin.



Jacinta Richardson wrote:
> Florentin wrote:
>> # Expect to print only a and b --> it prints all from x1 to x3 as above
>>  while( <>) {
>>    if(/$start/+1 ../$end/-1) {
>>        print ;
>>    }
>>  }
>
> After running through Perltidy, this is the same as writing:
>
> while (<>) {
>     if ( /$start/ + 1 .. /$end/ - 1 ) {
>         print;
>     }
> }
>
> which of course is the same as writing:
>
> while (<>) {
>     if ( ((/$start/) + 1) .. ((/$end/) - 1) ) {
>         print;
>     }
> }
>
> So, we're doing the regular expression (in a scalar context) and it's
> returning true or false then we're adding (or subtracting) one from
> that value. So to walk through you're input we get:
>
> # cat _a           # first regexp returns false, add 1 to make true
> # x1               # end of range is false
> # a                # end of range still false
> # x5               # second regexp returns true, minus 1 for a value
> of false
> # x1               # end of range still false
> # b                # end of range still false
> # x3               # second regexp: true minus 1, end of range still
> false
> # c                # end of range still false
> # d                # end of range still false
>
> so it prints everything.  Thanks to the plus one, the first part of
> your range is always true, and the minus one causes the end of the
> range to be false even when the regexp is true.
>
> I'm not quite sure what you wished to achieve with the above code, but
> that probably wasn't it.
>
>> # Expect to print only a and b --> prints nothing ??
>>  while( my $x = <>) {
>>       if($x=~m/$start/+1 .. $x=~m/$end/-1){
>>           print ;
>>       }
>>  }
>
> This behaves the same as the previous one.  Except you're printing $_
> rather than $x.  If you had turned warnings on for your test code,
> Perl would have told you about this.
>
> All the best,
>
>     Jacinta
>
>